Problem 53 - Thomas Calculus

Question

Determine how many terms should be used to estimate the sum of the entire series with an error of less than \(0.001 .\) $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n^{2}+3} $$

Step-by-Step Solution

Answer

Use 31 terms for an error less than 0.001.

Step 1: Understand the Alternating Series Test

The series in question is \( \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n^{2}+3} \). This is an alternating series, as indicated by the \((-1)^{n}\) term. The alternating series test states that if the absolute value term \( b_n = \frac{1}{n^2 + 3} \) is positive, decreasing, and its limit as \( n \to \infty \) is zero, the series converges.

Step 2: Apply the Alternating Series Approximation Error

For an alternating series where the terms decrease in absolute value, the error in approximating the sum by the first \( N \) terms is less than or equal to the absolute value of the first omitted term. That is, the error \( E_{N} \) satisfies \( |S - S_{N}| \leq |a_{N+1}| \). So, we want \( |a_{N+1}| = \left|\frac{1}{(N+1)^2 + 3}\right| < 0.001 \).

Step 3: Solve the Inequality for Error Less than 0.001

Set the inequality \( \frac{1}{(N+1)^2 + 3} < 0.001 \). To solve for \( N+1 \), begin by inverting the inequality: \((N+1)^2 + 3 > 1000 \). Thus, \( (N+1)^2 > 997 \). Next, solve for \( N+1 \) by taking the square root: \( N+1 > \sqrt{997} \approx 31.56 \). Therefore, \( N+1 \) must be at least 32.

Step 4: Determine the Required Number of Terms (N)

Since \( N+1 \) must be at least 32, the smallest integer \( N \) is 31 (since \( N+1 = 32 \) implies \( N = 31 \)). Therefore, you should use \( N = 31 \) terms to ensure the error is less than \( 0.001 \).

Key Concepts

Alternating Series Test

An alternating series is quite special due to its unique pattern of signs, alternating between positive and negative. This occurs in the form \((-1)^n a_n\), where \((-1)^n\) introduces the change in sign and \(a_n\) is a sequence of positive terms. A key feature of alternating series is that they can converge to a sum even when the series is infinite. To determine convergence, we use the Alternating Series Test.

The Alternating Series Test offers a straightforward set of conditions to check:

• The sequence \(b_n\) derived from the absolute values \(a_n = (-1)^n b_n\) of the series must be positive.
• It must be decreasing. In other words, each term must be smaller than, or equal to, the preceding term such as \(b_{n+1} \leq b_n\).
• Finally, as \(n\) becomes very large, the series terms must approach zero, \lim_{n \to \infty} b_n = 0\.

If these criteria are met, the alternating series converges, meaning it has a definite sum that can be estimated with a finite series.

Error Estimation

One practical advantage of alternating series lies in estimating the error in approximating the infinite series's sum by a finite number of terms. When using only a limited number of terms to approximate an infinite series, we inevitably introduce error.

For alternating series, the error \(E_N\) of truncating the series after \(N\) terms is bounded by the absolute value of the first omitted term. This means the error is less significant, closer to the value of the next term you didn't include, \(|S - S_N| \leq |a_{N+1}|\).

Thus, if we require the error to be less than a specified tolerance, say \(0.001\), we simply determine the smallest \(N\) for which \(|a_{N+1}| < 0.001\). This approach simplifies approximating the sum while managing the precision of the series.

Convergence of Series

Convergence is a central concept in series, signifying that as you add more and more terms, the total sum approaches a fixed number. Alternating series often converge if they fulfill the conditions of the Alternating Series Test.

In the example of \((-1)^n \frac{1}{n^2 + 3}\), after verifying that the absolute terms \(\frac{1}{n^2 + 3}\) are positive and decrease, and their limit approaches zero, we can confirm convergence. This means you can approximate the infinite series by a partial sum, with increasing accuracy as you include more terms.

Convergence doesn't guarantee rapid approximation. Sometimes a significant number of terms is needed to achieve desired precision, like when estimating with errors under \(0.001\). The series converging ensures that using sufficient terms will result in a sum indistinguishably close to the actual infinite sum.